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My understanding is that in a simple passive RC low pass filter, the high frequency noise will just go to ground while the low frequency signal will go through the resistor only. My question is what happens to the high frequency signal in the below active low pass filter?

enter image description here

The capacitor is not connected to ground, so where does the high frequency signal go?

I mean in theory, the inverting input is technically connected to ground since the op amp will do what it can to make the difference in voltage between both inputs to be zero, but what about if it was a differential amplifier?

in that case, the inverting input will not be at ground, so where does the high frequency signal go?

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  • \$\begingroup\$ Where does the high frequency signal go in an LR low pass filter? \$\endgroup\$
    – Andy aka
    Commented Nov 19, 2021 at 12:50

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The high-frequency current gets sunk¹ into the opamp's output (and finally conducted to the supply rails of the opamp).

It's the opamp's job here to actively cancel the noise, which you'll find is exactly what happens if you apply the golden opamp rules and write down where the current from the input goes.


¹ not quite the right word, because high-frequency current has a negative direction half of the time, it just "goes into and comes out of" the opamp output.

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  • \$\begingroup\$ until a specific frequency is reach and then the noise totally bypasses the OPAMP and goes via the capacitor \$\endgroup\$
    – user16222
    Commented Nov 19, 2021 at 11:38
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    \$\begingroup\$ let's assume ideal components with infinite bandwidth for a analytical problem, shall we ;) \$\endgroup\$ Commented Nov 19, 2021 at 11:39
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    \$\begingroup\$ fair enough :) im just use to my EMC consultants ripping apart my band-limited diff-amp (screwing up my PQ performance) because there is a +2dB in the 20MHz range :) \$\endgroup\$
    – user16222
    Commented Nov 19, 2021 at 12:35
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In this example circuit, two things happen : the textbook answer per Marcus's answer, and the practical answer.

The textbook answer applies to small signal HF signals within the bandwidth of the operational amplifier.

Large signals, or signals outside the opamp's bandwidth, don't behave in such an ideal manner.

First assume Vout=0, reflecting the limited open loop bandwidth and slew rate of the opamp. (This is an over-simplification; as JonRB points out, C1 can also deliver noise directly to the load)

Treat R1*C as a first order low pass filter, and calculate the resultant input voltage at In- of the opamp (given the input noise signal you have). If it is significant (tens or hundreds of mV) at frequencies where

  • (a) the opamp open loop gain is low,
  • (b) the input dV/dt * the open loop gain at that frequency exceeds the output slew rate

then the input stage will behave non-linearly adding such effects as demodulating the noise signal to DC (or LF reflecting its envelope), intermodulation with wanted input signals.

Given large HF input signals, I prefer a passive filter ahead of the opamp, or an adequately fast opamp.

I don't know how well SPICE models reflect nonlinear response to large HF input signals.

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what happens to the high frequency signal in the below active low pass filter?

enter image description here

Assuming the high frequency component under consideration is within the bandwidth of the op-amp, the output contains a high frequency component which approximately matches the high frequency component at the inverting input of the op-amp, except that it is approximately 180 degrees out of phase. When the input high frequency component and the output high frequency component meet at the inverting input of the op-amp, they nearly completely cancel, so the resultant, when amplified remains small. The low frequency component, on the other hand is amplified approximately. according to the ratio of R2/R1. So, relative to the low frequency component, the high frequency component is attenuated.

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